Table of Contents. Pg. # Momentum & Impulse (Bozemanscience Videos) 1 1/11/16

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1 Table of Contents g. # 1 1/11/16 Momentum & Impulse (Bozemanscience Videos) 2 1/13/16 Conservation of Momentum 3 1/19/16 Elastic and Inelastic Collisions 4 1/19/16 Lab 1 Momentum 5 1/26/16 Rotational Dynamics

2 Chapter 7 Momentum & Impulse In physics, work is done by forces on an object over a distance. 1. Impulse, J = F t (N s) 2. Momentum, p = m v (kg m/s) 3. Impulse-Momentum Theorem, F t = p = mvf mvi (kg m/s) 4. Conservation of Momentum, pf = pi (kg m/s) Force F t 0 t f Time t

3 hysics 1 Exam Description

4 Warm Up (1/25/16) Glancing Collision 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. fter the collision, the 0.20=kg puck has a speed of 1.0 m/s at an angle of θ = 53 to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision. = m 1 m 2 = m 1 = u 1 v 1 = v 2 =0 m 2 u 2 =?

5 Warm Up (1/25/16) Glancing Collision v 1 = = m 1 m 2 = v 2 =0 m 1 m 2 u 2 =? Elastic Collision (balls bounce off each other) 0 m 1 v 1 = m 1 u 1 Cosθ 1 + m 2 u 2 Cosθ 2 F net = ma = m v = mv = p 0 = m 1 u 1 inθ 1 m 2 u 2 inθ 2 ext t t t u 2 = m 1 u 1 inθ 1 0 = p p f = p i 0, at rest! m 2 inθ 2 t m 1 v 1 + m 2 v 2 = m 1 u 1 + m 2 u 2 0 = p m 1 u 1 Cosθ 1 = + m 2 u 2 Cosθ 2 m 1 v 1 m 1 v ^ ^ 1 x + ( 0 ) y = (m ^ 1 u 1 Cosθ 1 + m 2 u 2 Cosθ 2 ) x + (m 1 u 1 inθ 1 m 2 u 2 inθ 2 ) ^ y = u 1

6 Warm Up (1/25/16) Glancing Collision v 1 = = m 1 m 2 = v 2 =0 m 1 m 2 u 2 =? = u 1 m 1 v 1 = m 1 u 1 Cosθ 1 + m 2 u 2 Cosθ 2 0 = m 1 u 1 inθ 1 m 2 u 2 inθ 2 u 2 = m 1 u 1 inθ 1 m 2 inθ 2 m 1 u 1 Cosθ 1 = m 1 v 1 m 2 u 2 Cosθ 2 u 1 = m 1 v 1 m 2 m 1 u 1 inθ 1 Cosθ 2 m 2 inθ 2 m 1 Cosθ 1 Cotθ 2 = v 1 Cosθ 1 u 1 Tanθ 2 = u 1 Tanθ 1 u 1 Tanθ 1 v 1 Cosθ 1 u 1 θ 2 =Tan -1 [ 1m/s Tan53 (2m/s Cos53 1m/s ] = 29.7

7 Warm Up (1/26/16) Glancing Collision Object 1 is moving along the x axis with an initial momentum of +16 kg m/s, where the + sign indicates that it is moving to the right. s the drawing shows, object 1 collides with a second object that is initially at rest. The collision is not head-on, so the objects move off in different directions after the collision. The net external force acting on the two-object system is zero. fter the collision, object 1 has a momentum whose y component is 5 kg m/s. What is the y component of the momentum of object 2 after the collision? (a) 0 kg m/s (b) +16 kg m/s (c) +5 kg m/s (d) 16 kg m/s (e) The y component of the momentum of object 2 cannot be determined y 1 2 +x

8 Warm Up (1/26/16) Glancing Collision p f = p i 0, at rest! p 1i + p 2i = p 1f + p 2f (16 kg m/s) ^ x + ( 0 ) ^ y = (p 1f Cosθ 1 + p 2f Cosθ 2 ) ^ x + ( 5kg m/s + p 2yf )^y 0 = ( 5kg m/s + p 2yf ) p 2yf = 5kg m/s +y x 1

9 Warm Up (1/27/16) Explosion By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor. s the plate falls, its momentum has only a vertical component, and no component parallel to the floor. fter the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing, find the masses of pieces 1 and m/s 25.0 m 1 m 3 = 1.30 kg 1.79 m/s m m/s roblem 23

10 Warm Up (1/27/16) Explosion p i = p f ( 0 ) ^ x+ ( 0 ) ^y + (m = (-m 1 u 1 Cosθ 1 + m 2 u 2 Cosθ 2 ) ^x 1 + m 2 + m 3 )v ^z + u 1 = 3.00 m/s (m 1 u 1 inθ 1 + m 2 u 2 inθ 2 m 3 u 3 ) ^y + ( 0 ) ^z θ 1 = = ( m 1 u 1 Cosθ 1 + m 2 u 2 Cosθ 2 ) m 1 u 2 = 1.79 m/s m (m 1 + m 2 + m 3 )v = = θ 2 (m 1 u 1 inθ 1 + m 2 u 2 inθ 2 + m 3 u 3 ) m 3 = 1.30 kg 3.07 m/s = u 3

11 Warm Up (1/27/16) Explosion p i = p f ( 0 ) ^ x+ ( 0 ) ^y + (m 1 + m 2 + m 3 )v ^z= (-m 1 u 1 inθ 1 + m 2 u 2 Cosθ 2 ) ^x + (m ( 0 ) ^z 1 u 1 Cosθ 1 + m 2 u 2 inθ 2 m 3 u 3 ) ^y + m 2 = 1.58 kg 0 = ( m 1 u 1 inθ 1 + m 2 u 2 Cosθ 2 ) 0 = (m 1 u 1 Cosθ 1 + m 2 u 2 inθ 2 m 3 u 3 ) m 1 = m 3 u 3 u 1 (inθ 1 +Cosθ 1 ) (m 1 + m 2 + m 3 ) v = 0 0 = m 1 u 1 (inθ 1 + Cosθ 1 ) + m 2 u 2 (Cosθ 2 inθ 2 ) + m 3 u 3 m 2 = m 1 u 1 inθ 1 m 3 u 3 = m 1 u 1 (inθ 1 + Cosθ 1 ) u 2 Cosθ 2 0, in45 = Cos45 = 0.94(3m/s)in m/sCos45 = 1.30kg(3.07m/s) = 0.94 kg 3m/s(in45 +Cos45 )

12 Chapter 8 Rotational Kinematics Rolling motion involves both linear and rotational motion. If there is no slipping at the point of contact with the ground, the motion of the tire of an car is an example of rolling motion. While the car is moving with linear speed v covering a distance d, a point on the outer edge of the tire moves the same distance along the circular path s = θ r = d. Thus, the linear speed v = d t of the car is the same as the tangential speed of a point on the outer edge of the tire: v T = s t = θ r t = ( θ t) r = ω r r Linear velocity, v (a) B s B d = s

13 Chapter 8 Rotational Kinematics Vector Nature of ngular Variables Like linear velocity and acceleration, also angular velocity (ω) and angular acceleration (α) are vector quantities. They both have magnitude and direction ω = Δθ Δt = θ f θ i Δt radians second ω Right hand α = Δω Δt = ω f ω i Δt radians second 2 Right hand ω 8.16

14 Warm Up (1/28/16) Rotational Kinematics wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the angular speed of the wheel is 2.00 rad/s at t i = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00s?

15 Warm Up (1/28/16) Rotational Kinematics wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the angular speed of the wheel is 2.00 rad/s at t i = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00s? Given: α = 3.50 rad/s 2, ω i = 2 rad/s, Δt = 2 s, Δθ =? Rotational Motion ( constant) 0 t 1 2 ( 0 )t x 0t 1 2 t Linear Motion (a constant) (8.4) v v 0 at (2.4) 1 (8.6) 2 (v 0 v)t (2.7) (8.7) x v 0 t 1 2 at 2 (2.8) (8.8) v 2 v 2 0 2ax (2.9)

16 Warm Up (1/28/16) Rotational Kinematics wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the angular speed of the wheel is 2.00 rad/s at t i = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00s? Given: α = 3.50 rad/s 2, ω i = 2 rad/s, Δt = 2 s, Δθ =? θ = ω i t + ½αt 2 = 2rad/s(2s) + ½(3.5rad/s 2 )(2s) 2 = 11.0 rad 1 rev 11.0 rad = 1.75 rev 2π rad 1 rev = 2π rad ω f = ω i + αt= 2rad/s + (3.50rad/s 2 )(2s)= 9 rad/s

17 Rotational Kinematics n airplane propeller slows from an initial angular speed of 12.5 rev/s to a final angular speed of 5.00 rev/s. During this process, the propeller rotates through 21.0 revolutions. Find the angular acceleration of the propeller in radians per second squared, assuming its constant. Given: ω i = 12.5 rev/s, ω f = 5 rev/s, Δθ = 21.0 rev, α =? ω i = 12.5 rev/s, 2π rad 1 rev ω f = 5 rev/s, Δθ = 21.0 rev ( 2π rad rev )= 42π rad 2π rad 1 rev = 25π rad/s = 10π rad/s

18 Rotational Kinematics Given: ω i = 12.5 rev/s, ω f = 5 rev/s, Δθ = 21.0 rev, α =? Δθ = 21.0 rev ( 2π rad rev )= 42π rad ω i = 12.5 rev/s, 2π rad 1 rev ω f = 5 rev/s, 2π rad 1 rev = 25π rad/s = 10π rad/s ω 2 = ω i 2 + 2αΔθ α = ω 2 ω 2 i 2Δθ = (10π rad/s) 2 (25π rad/s) 2 2(42π rad) α = 6.25 π rad/s 2

19 Warm Up (1/29/16) The Earth rotates about an axis of rotation that goes through an axis that goes through the geographic poles. a) ow b) What fast is the is a angular person in speed an of Francisco the Earth? moving? angular speed, ω = 2π =2π x 1 h = 7.27 x 10-5 rad/s T 24 h 3600 s v = r ω = ( RECO 38 ) ω = 6.38 x 106 m (Cos 38 ) (7.27 x 10-5 rad/s) v = 365 m/s x1.09 yards x Football Field 1m v = 3.98 Football Field s 100 yards

20 ow fast are we moving? The Earth rotates about an axis of rotation that goes through an axis that goes through the geographic poles. c) What is the centripetal acceleration d) force? ssume of m this = 77 person? kg. angular speed, ω = 2π =2π x 1 h = 7.27 x 10-5 rad/s T 24 h 3600 s FN v = 365 m/s a c = v2 = v2 Fg ( RECO 38 ) r 2 (365 m/s) ac = 6.38 x 106 m (Cos 38 ) ac = m/s2 FC = Fg + FN = m ac,inward FC = 77 kg ( m/s2),inward = 2.04 N,inward